### Orthogonal and symmetric maps: Classification of orthogonal maps

### Three-dimensional orthogonal maps

For the study of three-dimensional orthogonal maps we will use the #(2\times2)#-matrices \[ D_{\varphi} = \matrix{\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi)& \cos(\varphi)} \] for certain #\varphi#. These are the rotation matrices of *Two-dimensional orthogonal maps*.

Let #V# be a #3#-dimensional inner product space and #L:V\to V# an orthogonal map.

- If #L# is special orthogonal, then it is a
**rotation**, that is to say, there is a line passing through the origin, such that each vector along that line is fixed by #L#, and the restriction of #L# to the plane through the origin perpendicular to that line is a rotation. There is an orthonormal basis #\alpha# of #V# and an angle #\varphi# such that \[L_\alpha = \matrix{1&0\\ 0 & D_{\varphi}}\] - If #L# is non-special orthogonal, then it is an
**improper rotation**, that is to say, there is a line passing through the origin which is the eigenspace of #L# corresponding to eigenvalue #-1#, and the restriction of #L# to the plane through the origin perpendicular to that line is a rotation. There is an orthonormal basis #\alpha# of #V# and an angle #\varphi# such that \[L_\alpha = \matrix{-1&0\\ 0 & D_{\varphi}}\]

In both cases, the line is called an **axis** of rotation and the plane a **plane of rotation** or **rotation plane**.

If #L# does not equal #I_V# or #-I_V#, then its axis and rotation plane are unique.

special orthogonal

An orthogonal map #L# is *special orthogonal* if #\det (L_A)=1# and *non-special orthogonal* if #\det (L_A)=-1#. To find the answer, we determine the determinant of #L_A#, which is equal to #\det(A)#. For this purpose we reduce #A# to *reduced echelon form*, as shown below. Here we record in the second column, by which number the newly obtained matrix has to be multiplied in order to get the determinant of the previous matrix. In the third column we record the product of these factors.

\[\begin{array}{c|r|r}

\text{matrix}&\text{step}&\text{total}\\ \hline \matrix{-{{18}\over{23}} & -{{3}\over{23}} & {{14}\over{23}} \\ -{{13}\over{23}} & -{{6}\over{23}} & -{{18}\over{23}} \\ -{{6}\over{23}} & {{22}\over{23}} & -{{3}\over{23}} \\ } & -1 & -1 \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ -{{13}\over{23}} & -{{6}\over{23}} & -{{18}\over{23}} \\ -{{6}\over{23}} & {{22}\over{23}} & -{{3}\over{23}} \\ } & -{{23}\over{18}} & {{23}\over{18}} \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ -{{6}\over{23}} & {{22}\over{23}} & -{{3}\over{23}} \\ } & 1 & {{23}\over{18}} \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ 0 & 1 & -{{1}\over{3}} \\ } & 1 & {{23}\over{18}} \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ } & -1 & -{{23}\over{18}} \\ \matrix{1 & 0 & -{{13}\over{18}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ } & 1 & -{{23}\over{18}} \\ \matrix{1 & 0 & -{{13}\over{18}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & 0 & -{{23}\over{18}} \\ } & 1 & -{{23}\over{18}} \\ \matrix{1 & 0 & -{{13}\over{18}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & 0 & 1 \\ } & -{{18}\over{23}} & 1 \\ \matrix{1 & 0 & 0 \\ 0 & 1 & -{{1}\over{3}} \\ 0 & 0 & 1 \\ } & 1 & 1 \\ \matrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ } & 1 & 1

\end{array}\] We conclude that the determinant of #A# equals #1# so the answer is: special orthogonal.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.